Class 10: Linear Equations – CBSE Board Questions

$\displaystyle \textbf{Question 1. }\text{The value of }p\text{ for which the equations }px+3y=p-3, \\ 12x+py=p\text{ has infinitely many solutions is:} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }-6\text{ only} \qquad \text{(b) }6\text{ only}$
$\displaystyle \text{(c) }\pm6 \qquad \text{(d) Any real number except }\pm6$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{1. }(b)\ \text{The given equations are}$
$\displaystyle px+3y-p+3=0$
$\displaystyle 12x+py-p=0$
$\displaystyle \text{For infinitely many solutions,}$
$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\displaystyle \Rightarrow \frac{p}{12}=\frac{3}{p}=\frac{-p+3}{-p}$
$\displaystyle \frac{p}{12}=\frac{3}{p}$
$\displaystyle \Rightarrow p^{2}=36$
$\displaystyle \Rightarrow p=\pm6$
$\displaystyle \frac{3}{p}=\frac{-p+3}{-p}$
$\displaystyle \Rightarrow -3p=p^{2}-3p$
$\displaystyle \Rightarrow p^{2}-6p=0$
$\displaystyle \Rightarrow p(p-6)=0$
$\displaystyle \Rightarrow p=0,\ p=6$
$\displaystyle \therefore p=6\text{ only}$

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$\displaystyle \textbf{Question 2. }\text{In the given figure, graphs of two linear equations are shown. The pair} \\ \text{of these linear equations is:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) consistent with unique solution.}$
$\displaystyle \text{(b) consistent with infinitely many solutions.}$
$\displaystyle \text{(c) inconsistent.}$
$\displaystyle \text{(d) inconsistent but can be made consistent by} \\ \text{extending these lines.}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{2. }(d)\ \text{Inconsistent but can be made consistent by extending these lines.}$

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$\displaystyle \textbf{Question 3. }\text{The pair of equations }x=2a\text{ and }y=3b\ (a,b\neq0)\text{ graphically} \\ \text{represents straight lines which are:} \hspace{0.2cm}\text{[CBSE 2024(C)]}$
$\displaystyle \text{(a) coincident}$
$\displaystyle \text{(b) parallel}$
$\displaystyle \text{(c) intersecting at }(2a,3b)$
$\displaystyle \text{(d) intersecting at }(3b,2a)$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{3. }(c)\ \text{Intersecting at }(2a,3b)$

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$\displaystyle \textbf{Question 4. }\text{The pair of linear equations }2x=5y+6\text{ and }15y=6x-18\text{ represents} \\ \text{two lines which are:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) intersecting}$
$\displaystyle \text{(b) parallel}$
$\displaystyle \text{(c) coincident}$
$\displaystyle \text{(d) either intersecting or parallel}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{4. }(c)\ \text{Given equations are }2x-5y-6=0\text{ and }6x-15y-18=0$
$\displaystyle \text{Here, }\frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3},\ \frac{b_{1}}{b_{2}}=\frac{-5}{-15}=\frac{1}{3}$
$\displaystyle \frac{c_{1}}{c_{2}}=\frac{-6}{-18}=\frac{1}{3}$
$\displaystyle \Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\displaystyle \therefore \text{Lines are coincident.}$

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$\displaystyle \textbf{Question 5. }\text{Graphically, the pair of linear equations }3x-y+8=0\text{ and } \\ 3x-y=24\text{ represents two lines which are:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{(a) intersecting exactly at one point}$
$\displaystyle \text{(b) intersecting exactly at two points}$
$\displaystyle \text{(c) coincident}$
$\displaystyle \text{(d) parallel}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{5. }(d)\ \text{The given equations are:}$
$\displaystyle 3x-y+8=0\text{ and }3x-y-24=0$
$\displaystyle \text{Comparing with }a_{1}x+b_{1}y+c_{1}=0\text{ and }a_{2}x+b_{2}y+c_{2}=0,$
$\displaystyle a_{1}=3,\ b_{1}=-1,\ c_{1}=8$
$\displaystyle a_{2}=3,\ b_{2}=-1,\ c_{2}=-24$
$\displaystyle \text{Now, }\frac{a_{1}}{a_{2}}=1,\ \frac{b_{1}}{b_{2}}=1,\ \frac{c_{1}}{c_{2}}=-\frac{1}{3}$
$\displaystyle \text{Clearly, }\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$
$\displaystyle \therefore \text{The given system represents parallel lines graphically.}$

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$\displaystyle \textbf{Question 6. }\text{Two lines are given to be parallel. The equation of one of the lines} \\ \text{is }3x-2y=5,\text{ the equation of the second line can be} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }9x+8y=7 \qquad \text{(b) }-12x-8y=7$
$\displaystyle \text{(c) }-12x+8y=7 \qquad \text{(d) }12x+8y=7$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{6. }(c)\ \text{As we know, lines }a_{1}x+b_{1}y+c_{1}=0\text{ and }a_{2}x+b_{2}y+c_{2}=0$
$\displaystyle \text{represent parallel lines if }\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$
$\displaystyle \text{So, equation of second line is }-12x+8y=7$
$\displaystyle \text{as }\frac{3}{-12}=\frac{-2}{8}\neq\frac{-5}{7}$

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$\displaystyle \textbf{Question 7. }\text{The value of }k\text{ for which the system of equations } \\ x+y-4=0\text{ and }2x+ky=3,\text{ has no solution, is} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{(a) }-2 \qquad \text{(b) }\neq2$
$\displaystyle \text{(c) }3 \qquad \text{(d) }2$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{7. }(d)\ \text{For no solution}$
$\displaystyle \frac{1}{2}=\frac{1}{k}\neq\frac{-4}{-3}$
$\displaystyle \Rightarrow k=2$

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$\displaystyle \textbf{Question 8. }\text{For what value of }k,\text{ the following system of equations } \\ kx+2y=3,\ 3x+6y=10\text{ has a unique solution.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{8. }\text{For unique solution, }\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}$
$\displaystyle \Rightarrow \frac{3}{3}\neq\frac{2}{6}$
$\displaystyle \Rightarrow 1\neq k$
$\displaystyle \therefore \text{The given system has a unique solution for all values of }k\text{ other than }1.$

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$\displaystyle \textbf{Question 9. }\text{Find }c\text{ if the system of equations }cx+3y+(3-c)=0; \\ 12x+cy-c=0\text{ has infinitely many solutions?} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{9. }\text{The system of equations is}$
$\displaystyle cx+3y+(3-c)=0 \qquad (i)$
$\displaystyle 12x+cy-c=0 \qquad (ii)$
$\displaystyle \text{For infinitely many solutions, we have}$
$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\displaystyle \Rightarrow \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}$
$\displaystyle \Rightarrow c^{2}=36\text{ and }-3c=3c-c^{2}$
$\displaystyle \Rightarrow c=\pm6\text{ and }c^{2}=6c$
$\displaystyle \Rightarrow c=\pm6\text{ and }c=0,6$
$\displaystyle \therefore \text{Possible value of }c=6.$

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$\displaystyle \textbf{Question 10. }\text{Find the value of }k\text{ for which the following pair of linear equations} \\ \text{have infinitely many solutions:} 2x+3y=7,\\ (k+1)x+(2k-1)y=4k+1. \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{10. }\text{For infinitely many solutions:}$
$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\displaystyle \Rightarrow \frac{2}{k+1}=\frac{3}{2k-1}=\frac{7}{4k+1}$
$\displaystyle \Rightarrow 2(2k-1)=3(k+1)$
$\displaystyle \text{and}$
$\displaystyle 3(4k+1)=7(2k-1)$
$\displaystyle \Rightarrow 4k-2=3k+3$
$\displaystyle \text{and}$
$\displaystyle 12k+3=14k-7$
$\displaystyle \Rightarrow k=5\text{ and }k=5$
$\displaystyle \therefore \text{Hence, the value of }k\text{ is }5.$

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$\displaystyle \textbf{Question 11. }\text{Solve the following pair of equations graphically:}$
$\displaystyle x+3y=6,\ 2x-3y=12$
$\displaystyle \text{Also, find the area of the triangle formed by the lines representing the equations} \\ \text{with }y\text{-axis.} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{11. }\text{Consider }x+3y=6.$
$\displaystyle \text{When }x=0,\ y=2$
$\displaystyle \text{When }x=6,\ y=0$

$\displaystyle \begin{array}{|c|c|c|}\hline x&0&6\\ \hline y&2&0\\ \hline \end{array}$

$\displaystyle \text{Consider }2x-3y=12.$
$\displaystyle \text{When }x=0,\ y=-4$
$\displaystyle \text{When }x=6,\ y=0$

$\displaystyle \begin{array}{|c|c|c|}\hline x&0&6\\ \hline y&-4&0\\ \hline \end{array}$ $\displaystyle \text{The required }\triangle\text{ is }\triangle ABC\text{ with vertices at }A(6,0),\ B(0,2)\text{ and }C(0,-4).$
$\displaystyle \text{Now, }\mathrm{ar}(\triangle ABC)=\frac{1}{2}\times BC\times OA$
$\displaystyle =\frac{1}{2}\times6\times6$
$\displaystyle =18\text{ sq units}$

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$\displaystyle \textbf{Question 12. }\text{Determine graphically the coordinates of the vertices of a triangle, the} \\ \text{equations of whose sides are given by}$
$\displaystyle 2y-x=8,\ 5y-x=14\text{ and }y-2x=1. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{12. }\text{Consider equation }2y-x=8$
$\displaystyle \Rightarrow x=2y-8$
$\displaystyle \text{Some points on graph are}$

$\displaystyle \begin{array}{|c|c|c|c|}\hline x&0&-8&-4\\ \hline y&4&0&2\\ \hline \end{array}$

$\displaystyle \text{Consider equation }5y-x=14$
$\displaystyle \Rightarrow x=5y-14$
$\displaystyle \text{Some points on graph are}$

$\displaystyle \begin{array}{|c|c|c|c|}\hline x&-4&1&6\\ \hline y&2&3&4\\ \hline \end{array}$

$\displaystyle \text{Consider equation }y-2x=1$
$\displaystyle \Rightarrow y=2x+1$
$\displaystyle \text{Some points on graph are}$

$\displaystyle \begin{array}{|c|c|c|c|}\hline x&0&1&2\\ \hline y&1&3&5\\ \hline \end{array}$

$\displaystyle \text{Plotting the points on graph, we get triangle }ABC\text{ with vertices }A(-4,2),\ B(1,3),\ C(2,5).$

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$\displaystyle \textbf{Question 13. }\text{For what values of }m\text{ and }n,\text{ the following system of} \\ \text{linear equations has infinitely many solutions.}$
$\displaystyle 3x+4y=12 \ \ (m+n)x+2(m-n)y=5m-1 \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{13. }\text{Given equations are}$
$\displaystyle 3x+4y=12 \qquad (i)$
$\displaystyle (m+n)x+2(m-n)y=5m-1 \qquad (ii)$
$\displaystyle \text{System of linear equations has infinitely many solutions if}$
$\displaystyle \frac{3}{m+n}=\frac{4}{2(m-n)}=\frac{12}{5m-1}$
$\displaystyle \text{Taking }\frac{3}{m+n}=\frac{4}{2(m-n)}$
$\displaystyle \Rightarrow 3(m-n)=2(m+n)$
$\displaystyle \Rightarrow m=5n \qquad (iii)$
$\displaystyle \text{Taking }\frac{4}{2(m-n)}=\frac{12}{5m-1}$
$\displaystyle \Rightarrow 2(5m-1)=12(m-n)$
$\displaystyle \Rightarrow 2m=12n-2 \qquad (iv)$
$\displaystyle \text{Put }m=5n\text{ in }(iv),\ \text{we get}$
$\displaystyle \Rightarrow 2(5n)=12n-2$
$\displaystyle \Rightarrow n=1$
$\displaystyle \therefore m=5(1)=5$

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$\displaystyle \textbf{Question 14. }\text{Draw the graphs of the pair of equations:}$
$\displaystyle x+2y=5\text{ and }2x-3y=-4. \\ \text{Also find the points where the lines meet the } x\text{-axis.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{14. }\text{Given equations are}$
$\displaystyle x+2y=5$
$\displaystyle \text{and}$
$\displaystyle 2x-3y=-4$
$\displaystyle \text{Table for }x+2y=5$

$\displaystyle \begin{array}{|c|c|c|c|c|}\hline x&5&3&7&-1\\ \hline y&0&1&-1&3\\ \hline \end{array}$

$\displaystyle \text{Plot values of }x\text{ and }y\text{ from above table on graph.}$

$\displaystyle \text{Table for }2x-3y=-4$

$\displaystyle \begin{array}{|c|c|c|c|c|}\hline x&-2&1&-5&4\\ \hline y&0&2&-2&4\\ \hline \end{array}$

$\displaystyle \text{Plot values of }x\text{ and }y\text{ from above table on graph.}$
$\displaystyle \text{The line }x+2y=5\text{ meets the }x\text{-axis at }(5,0)$
$\displaystyle \text{and the line }2x-3y=-4\text{ meets the }x\text{-axis at }(-2,0).$

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$\displaystyle \textbf{Question 15. }\text{The age of the father is twice the sum of the ages of his two children.} \\ \text{After }20\text{ years, his age will be} \text{ equal to the sum of the ages of his children. Find the} \\ \text{present age of the father.} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{15. }\text{Let the sum of present ages of }2\text{ children}=x\text{ years}$
$\displaystyle \text{Let the present age of father}=y\text{ years}$
$\displaystyle \text{According to first condition:}$
$\displaystyle y=2x \qquad (i)$
$\displaystyle \text{After }20\text{ years:}$
$\displaystyle \text{Sum of ages of }2\text{ children}=(x+40)\text{ years}$
$\displaystyle \text{Father's age}=(y+20)\text{ years}$
$\displaystyle \text{According to second condition:}$
$\displaystyle y+20=x+40$
$\displaystyle \Rightarrow y=x+20 \qquad (ii)$
$\displaystyle \text{Substituting the value of }y\text{ from }(i)\text{ in }(ii),$
$\displaystyle \Rightarrow 2x=x+20$
$\displaystyle \Rightarrow x=20$
$\displaystyle \text{Now, from }(i),$
$\displaystyle y=2\times20=40$
$\displaystyle \therefore \text{Present age of father}=40\text{ years}$

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$\displaystyle \textbf{Question 16. }\text{A man can row a boat downstream }20\text{ km in }2\text{ hours and upstream }$
$\displaystyle 4\text{ km in }2\text{ hours. Find }\text{his speed of rowing in still water. Also find the speed} \\ \text{of the stream.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{16. }\text{Let speed of the boat in still water be }x\text{ km/hr}$
$\displaystyle \text{and speed of stream be }y\text{ km/hr}$
$\displaystyle \text{Speed of boat upstream}=(x-y)\text{ km/hr}$
$\displaystyle \text{Speed of boat downstream}=(x+y)\text{ km/hr}$
$\displaystyle \text{According to given condition,}$
$\displaystyle \frac{20}{x+y}=2$
$\displaystyle \Rightarrow x+y=10 \qquad (i)$
$\displaystyle \text{and}$
$\displaystyle \frac{4}{x-y}=2$
$\displaystyle \Rightarrow x-y=2 \qquad (ii)$
$\displaystyle \text{From }(i),\ \text{we get}$
$\displaystyle y=10-x \qquad (iii)$
$\displaystyle \text{Substitute the value of }y\text{ in }(ii),\ \text{we get}$
$\displaystyle x-(10-x)=2$
$\displaystyle \Rightarrow 2x=12$
$\displaystyle \Rightarrow x=6$
$\displaystyle \text{Put }x=6\text{ in }(iii),\ \text{we get}$
$\displaystyle y=10-6=4$
$\displaystyle \therefore \text{Speed of boat in still water}=6\text{ km/hr}$
$\displaystyle \therefore \text{Speed of stream}=4\text{ km/hr}$

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$\displaystyle \textbf{Question 17. }\text{The values of }x\text{ and }y\text{ satisfying the two equations } \\ 32x+33y=34,\ 33x+32y=31\text{ respectively are:} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }-1,2 \qquad \text{(b) }-1,4 \qquad \text{(c) }1,-2 \qquad \text{(d) }-1,-4$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{17. }(a)\ \begin{cases}32x+33y=34 \qquad (i)\\[4pt]33x+32y=31 \qquad (ii)\end{cases}$
$\displaystyle \text{Adding }(i)\text{ and }(ii),\ \text{we get}$
$\displaystyle x+y=1 \qquad (iii)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),\ \text{we get}$
$\displaystyle x-y=-3 \qquad (iv)$
$\displaystyle \text{Adding }(iii)\text{ and }(iv),\ \text{we get }x=-1$
$\displaystyle \text{From }(iii),\ \text{we get }y=2$

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$\displaystyle \textbf{Question 18. }\text{Solve the following system of equations algebraically:}$
$\displaystyle 30x+44y=10;\ 40x+55y=13 \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{18. }\text{The given two equations are}$
$\displaystyle 30x+44y=10 \qquad (1)$
$\displaystyle 40x+55y=13 \qquad (2)$
$\displaystyle \text{Multiplying equation }(1)\text{ by }4\text{ and equation }(2)\text{ by }3\text{ and subtracting, we get}$
$\displaystyle 120x+176y=40$
$\displaystyle 120x+165y=39$
$\displaystyle \therefore 11y=1$
$\displaystyle \Rightarrow y=\frac{1}{11}$
$\displaystyle \text{Put the value of }y\text{ in equation }(1),\ \text{we get}$
$\displaystyle 30x+44\left(\frac{1}{11}\right)=10$
$\displaystyle 30x=6$
$\displaystyle x=\frac{6}{30}=\frac{1}{5}$
$\displaystyle \therefore x=\frac{1}{5},\ y=\frac{1}{11}$

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$\displaystyle \textbf{Question 19. }\text{Solve the following system of linear equations }7x-2y=5\text{ and } \\ 8x+7y=15\text{ and verify your answer.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{19. }\text{The given system of equations is}$
$\displaystyle 7x-2y=5 \qquad (i)$
$\displaystyle 8x+7y=15 \qquad (ii)$
$\displaystyle \text{On solving }(i)\text{ and }(ii),\ \text{we get }x=1,\ y=1$
$\displaystyle \text{Verification:}$
$\displaystyle 7x-2y=7\times1-2\times1=5$
$\displaystyle 8x+7y=8\times1+7\times1=15$
$\displaystyle \therefore \text{Hence, verified.}$

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$\displaystyle \textbf{Question 20. }\text{In Fig., ABCD is a rectangle. Find the values of }x\text{ and }y. \hspace{0.2cm}\text{[CBSE 2019]}$ $\displaystyle \text{Answer:}$

$\displaystyle \textbf{20. }\text{We know that opposite sides of a rectangle are equal.}$
$\displaystyle \therefore x+y=30 \qquad (i)$
$\displaystyle x-y=14 \qquad (ii)$
$\displaystyle \text{On adding,}$
$\displaystyle 2x=44$
$\displaystyle \therefore x=\frac{44}{2}=22$
$\displaystyle \text{Putting }x=22\text{ in equation }(i),\ \text{we have}$
$\displaystyle 22+y=30$
$\displaystyle \Rightarrow y=30-22=8$
$\displaystyle \therefore x=22\text{ cm and }y=8\text{ cm}$

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$\displaystyle \textbf{Question 21. }\text{If }x=a,\ y=b\text{ is the solution of the pair of equations } \\ x-y=2\text{ and }x+y=4,\text{ find the values of }a\text{ and }b. \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{21. }\text{Let }x=a\text{ and }y=b\text{ be the solution of the pair of linear equations}$
$\displaystyle x-y=2\text{ and }x+y=4,\ \text{so they must satisfy the equations.}$
$\displaystyle \therefore a-b=2 \qquad (i)$
$\displaystyle a+b=4 \qquad (ii)$
$\displaystyle \text{On adding,}$
$\displaystyle 2a=6$
$\displaystyle \Rightarrow a=3$
$\displaystyle \therefore b=a-2=3-2=1$

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$\displaystyle \textbf{Question 22. }\text{Solve the following pair of linear equations:}$
$\displaystyle y-4x=1$
$\displaystyle 6x-5y=9 \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{22. }\begin{cases}-4x+y=1 \qquad (i)\\[4pt]6x-5y=9 \qquad (ii)\end{cases}$
$\displaystyle \text{Operate: }5\times\text{equation }(i)+\text{equation }(ii)$
$\displaystyle -20x+5y=5$
$\displaystyle 6x-5y=9$
$\displaystyle \text{On adding,}$
$\displaystyle -14x=14$
$\displaystyle \therefore x=\frac{14}{-14}=-1$
$\displaystyle \text{Putting value of }x=-1\text{ in equation }(i),\ \text{we get}$
$\displaystyle -4(-1)+y=1$
$\displaystyle \Rightarrow y=-3$

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$\displaystyle \textbf{Question 23. }\text{The monthly incomes of two persons are in the ratio }9:7\text{ and }$
$\displaystyle \text{their monthly expenditures are in the ratio }4:3.\text{ If each saved Rs }5,000,\text{ express }$
$\displaystyle \text{the given situation algebraically as a system of linear equations in two variables.}$

$\displaystyle \text{Hence, find their respective monthly incomes.} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{23. }\text{Monthly incomes of two persons }A\text{ and }B\text{ are in the ratio }9:7.$
$\displaystyle \text{Monthly expenditures are in the ratio }4:3.$
$\displaystyle \text{Each saves Rs }5000.$
$\displaystyle \text{As, Total income = Expenditure + Saving}$
$\displaystyle \text{Let monthly income of }A=9x,\ \text{and }B=7x$
$\displaystyle \text{Monthly expenditure of }A=4y,\ \text{and }B=3y$
$\displaystyle \text{In the form of linear equations, we write}$
$\displaystyle 9x=4y+5000$
$\displaystyle 7x=3y+5000$
$\displaystyle \Rightarrow 9x-4y-5000=0 \qquad (1)$
$\displaystyle \Rightarrow 7x-3y-5000=0 \qquad (2)$
$\displaystyle \text{Here, we have to find out the value of }x.$
$\displaystyle \text{Multiplying equation }(1)\text{ by }3\text{ and equation }(2)\text{ by }4\text{ and subtracting, we get}$
$\displaystyle 27x-12y-15000=0$
$\displaystyle 28x-12y-20000=0$
$\displaystyle \therefore -x+5000=0$
$\displaystyle \Rightarrow x=5000$
$\displaystyle \therefore \text{Monthly income of }A=9\times5000=\text{Rs }45000$
$\displaystyle \therefore \text{Monthly income of }B=7\times5000=\text{Rs }35000$

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$\displaystyle \textbf{Question 24. }\text{Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi }$
$\displaystyle \text{will be twice as old as Nazma. How old are Rashmi and Nazma now?} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{24. }\text{Let the present age of Rashmi be }x\text{ years}$
$\displaystyle \text{and the present age of Nazma be }y\text{ years.}$
$\displaystyle 3\text{ years ago:}$
$\displaystyle \text{Rashmi's age}=(x-3)\text{ years}$
$\displaystyle \text{Nazma's age}=(y-3)\text{ years}$
$\displaystyle \text{According to first condition, }x-3=3(y-3)$
$\displaystyle \Rightarrow x-3=3y-9$
$\displaystyle \Rightarrow x-3y=-6 \qquad (i)$
$\displaystyle 10\text{ years later:}$
$\displaystyle \text{Rashmi's age}=(x+10)\text{ years}$
$\displaystyle \text{Nazma's age}=(y+10)\text{ years}$
$\displaystyle \text{According to second condition, }x+10=2(y+10)$
$\displaystyle \Rightarrow x+10=2y+20$
$\displaystyle \Rightarrow x-2y=10 \qquad (ii)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),\ \text{we get}$
$\displaystyle (x-2y)-(x-3y)=10-(-6)$
$\displaystyle \Rightarrow y=16$
$\displaystyle \text{Substituting the value of }y\text{ in }(i),\ \text{we get}$
$\displaystyle x-3\times16=-6$
$\displaystyle \Rightarrow x=42$
$\displaystyle \text{So, Rashmi's present age is }42\text{ years}$
$\displaystyle \text{and Nazma's present age is }16\text{ years.}$

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$\displaystyle \textbf{Question 25. }\text{A fraction becomes }\frac{1}{3}\text{ when }2\text{ is subtracted from the numerator}$
$\displaystyle \text{and it becomes } \frac{1}{2}\text{ when }1\text{ is subtracted} \text{from the denominator. Find the fraction.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{25. }\text{Let the numerator be }x\text{ and denominator be }y.$
$\displaystyle \text{As per first condition,}$
$\displaystyle \frac{x-2}{y}=\frac{1}{3}$
$\displaystyle \Rightarrow 3x-6=y$
$\displaystyle \Rightarrow 3x-y=6 \qquad (i)$
$\displaystyle \text{As per second condition,}$
$\displaystyle \frac{x}{y-1}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x=y-1$
$\displaystyle \Rightarrow 2x-y=-1 \qquad (ii)$
$\displaystyle \text{Equation }(i)-\text{equation }(ii),$
$\displaystyle x=7$
$\displaystyle \text{and }y=2x+1=2\times7+1$
$\displaystyle \Rightarrow y=14+1=15$
$\displaystyle \therefore \text{Fraction is }\frac{7}{15}.$

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$\displaystyle \textbf{Question 26. }\text{A part of monthly Hostel charge is fixed and the remaining depends on the number}$
$\displaystyle \text{of days one has taken food in the mess. When Swati takes food for }20\text{ days, she has to pay}$
$\displaystyle \text{Rs 3000 as hostel charges, whereas Mansi takes food for }25\text{ days and pays Rs }3500$
$\displaystyle \text{ as hostel charges. Find the fixed charges and the cost of food per day.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{26. }\text{Let fixed hostel charges be Rs }x.$
$\displaystyle \text{Let charges for food per day be Rs }y.$
$\displaystyle \text{Charges for food paid by Swati}=\text{Rs }3000$
$\displaystyle \therefore \text{First condition is }x+20y=3000 \qquad (i)$
$\displaystyle \text{Charges paid by Mansi}=\text{Rs }3500$
$\displaystyle \therefore \text{Second condition is }x+25y=3500 \qquad (ii)$
$\displaystyle \text{Subtracting }(ii)\text{ from }(i),$
$\displaystyle x+20y=3000$
$\displaystyle x+25y=3500$
$\displaystyle \therefore -5y=-500$
$\displaystyle \Rightarrow y=\text{Rs }100$
$\displaystyle \text{Putting value of }y=100\text{ in equation }(i),\ \text{we get}$
$\displaystyle x+20(100)=3000$
$\displaystyle x=3000-2000=\text{Rs }1000$
$\displaystyle \text{Hence, fixed hostel charges are Rs }1000\text{ and charges for food per day is Rs }100.$

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$\displaystyle \textbf{Question 27. }\text{Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey Rs }$
$\displaystyle x\text{ per student } \text{and Cricket Rs }y\text{ per student. School 'P' decided to award a total of Rs }$
$\displaystyle 9,500\text{ for the two games to }5\text{ and } 4\text{ students respectively; while school 'Q' decided}$
$\displaystyle \text{to award Rs } 7,370\text{ for the two games to }4\text{ and }3\text{ students} \text{ respectively.} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Based on the above information, answer the following questions:}$
$\displaystyle \text{(i) Represent the following information algebraically (in terms of }x\text{ and }y\text{).}$
$\displaystyle \text{(ii) What is the prize amount for hockey?}$
$\displaystyle \text{OR}$
$\displaystyle \text{Prize amount on which game is more and by how much?}$
$\displaystyle \text{(iii) What will be the total prize amount if there are }2\text{ students each from two games?}$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{27. }(i)\ \text{For school P}$
$\displaystyle 5x+4y=9500$
$\displaystyle \text{For school Q}$
$\displaystyle 4x+3y=7370$
$\displaystyle \text{(ii) Solving }5x+4y=9500\text{ and }4x+3y=7370,\ \text{we get}$
$\displaystyle x=980,\ y=1150$
$\displaystyle \therefore \text{Prize amount of Hockey}=\text{Rs }980$
$\displaystyle \text{OR}$
$\displaystyle \text{Prize amount for Hockey is Rs }980\text{/student}$
$\displaystyle \text{and prize amount for cricket is Rs }1150\text{/student.}$
$\displaystyle \text{So, prize amount on cricket is more than hockey by Rs }(1150-980)\text{ i.e. Rs }170.$
$\displaystyle \text{(iii) Total prize amount if there are }2\text{ students of each game}$
$\displaystyle =2\times980+2\times1150=\text{Rs }4260$

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$\displaystyle \textbf{Question 28. }\text{A book store shopkeeper gives books on rent for reading. He has variety }$
$\displaystyle \text{of books in his store related to fiction, stories and quizzes etc. He takes a fixed charge}$
$\displaystyle \text{for the first two days and an additional charge for subsequent day. Amruta paid Rs }$
$\displaystyle 22\text{ for a book and kept for }6\text{ days; while Radhika paid } \text{Rs }16\text{ for keeping the book for }4\text{ days.}$
$\displaystyle \text{Assume that the fixed charge be Rs }x\text{ and additional charge (per day) be Rs }y. \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{Based on the above information, answer any four of the following questions:}$
$\displaystyle \text{(i) The situation of amount paid by Radhika, is algebraically represented by}$
$\displaystyle \text{(a) }x-4y=16 \qquad \text{(b) }x+4y=16$
$\displaystyle \text{(c) }x-2y=16 \qquad \text{(d) }x+2y=16$
$\displaystyle \text{(ii) The situation of amount paid by Amruta, is algebraically represented by}$
$\displaystyle \text{(a) }x-2y=11 \qquad \text{(b) }x-2y=22$
$\displaystyle \text{(c) }x+4y=22 \qquad \text{(d) }x-4y=11$
$\displaystyle \text{(iii) What are the fixed charges for a book?}$
$\displaystyle \text{(a) Rs }9 \qquad \text{(b) Rs }10 \qquad \text{(c) Rs }13 \qquad \text{(d) Rs }15$
$\displaystyle \text{(iv) What are the additional charges for each subsequent day for a book?}$
$\displaystyle \text{(a) Rs }6 \qquad \text{(b) Rs }5 \qquad \text{(c) Rs }4 \qquad \text{(d) Rs }3$
$\displaystyle \text{(v) What is the total amount paid by both, if both of them have kept the book for }2\text{ more days?}$
$\displaystyle \text{(a) Rs }35 \qquad \text{(b) Rs }52 \qquad \text{(c) Rs }50 \qquad \text{(d) Rs }58$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{28. }(i)\ (d)\ \text{Fixed charges for first }2\text{ days}=\text{Rs }x$
$\displaystyle \text{Additional charge per day}=\text{Rs }y$
$\displaystyle \text{Radhika kept book for }4\text{ days and paid Rs }16$
$\displaystyle \text{ATQ, }x+2y=16$
$\displaystyle \text{(ii) }(c)\ \text{Amruta kept book for }6\text{ days and paid Rs }22$
$\displaystyle \text{ATQ, }x+4y=22$
$\displaystyle \text{(iii) }(b)\ \text{For Radhika, }x+2y=16 \qquad (1)$
$\displaystyle \text{For Amruta, }x+4y=22 \qquad (2)$
$\displaystyle \text{Solving }(1)\text{ and }(2),\ \text{we get}$
$\displaystyle x=10,\ y=3$
$\displaystyle \therefore \text{Fixed charges for book}=\text{Rs }10$
$\displaystyle \text{(iv) }(d)\ \text{Additional charges per day}=y=\text{Rs }3$
$\displaystyle \text{(v) }(c)\ \text{Total amount paid by both}$
$\displaystyle =(x+4y)+(x+6y)$
$\displaystyle =2x+10y$
$\displaystyle =2\times10+10\times3$
$\displaystyle =\text{Rs }50$